OFFSET
1,1
COMMENTS
For similar sequences that alternate in descending blocks of k natural numbers, we have: a(n) = (-1)^floor(n/k) * Sum_{j=1..k-1} (floor((n-j)/k) - floor((n-j-1)/k)) * (Product_{i=1..j} n-i-j+k+1) + Sum_{j=1..n} (-1)^(floor(j/k)+1) * (floor(j/k) - floor((j-1)/k)) * (Product_{i=1..k} j-i+1). Here, k=3.
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,-4,4,0,-6,6,0,-4,4,0,-1,1).
FORMULA
From Colin Barker, Oct 01 2018: (Start)
G.f.: 3*x*(1 + x + 2*x^3 - 4*x^4 - 30*x^5 + x^6 - 5*x^7 + 24*x^8) / ((1 - x)*(1 + x)^4*(1 - x + x^2)^4).
a(n) = a(n-1) - 4*a(n-3) + 4*a(n-4) - 6*a(n-6) + 6*a(n-7) - 4*a(n-9) + 4*a(n-10) - a(n-12) + a(n-13) for n>13.
(End)
EXAMPLE
a(1) = 3;
a(2) = 3*2 = 6;
a(3) = 3*2*1 = 6;
a(4) = 3*2*1 - 6 = 0;
a(5) = 3*2*1 - 6*5 = -24;
a(6) = 3*2*1 - 6*5*4 = -114;
a(7) = 3*2*1 - 6*5*4 + 9 = -105;
a(8) = 3*2*1 - 6*5*4 + 9*8 = -42;
a(9) = 3*2*1 - 6*5*4 + 9*8*7 = 390;
a(10) = 3*2*1 - 6*5*4 + 9*8*7 - 12 = 378;
a(11) = 3*2*1 - 6*5*4 + 9*8*7 - 12*11 = 258;
a(12) = 3*2*1 - 6*5*4 + 9*8*7 - 12*11*10 = -930;
a(13) = 3*2*1 - 6*5*4 + 9*8*7 - 12*11*10 + 15 = -915;
a(14) = 3*2*1 - 6*5*4 + 9*8*7 - 12*11*10 + 15*14 = -720;
a(15) = 3*2*1 - 6*5*4 + 9*8*7 - 12*11*10 + 15*14*13 = 1800;
a(16) = 3*2*1 - 6*5*4 + 9*8*7 - 12*11*10 + 15*14*13 - 18 = 1782;
a(17) = 3*2*1 - 6*5*4 + 9*8*7 - 12*11*10 + 15*14*13 - 18*17 = 1494;
a(18) = 3*2*1 - 6*5*4 + 9*8*7 - 12*11*10 + 15*14*13 - 18*17*16 = -3096;
etc.
MAPLE
a:=(n, k)->(-1)^(floor(n/k))* add((floor((n-j)/k)-floor((n-j-1)/k))*(mul(n-i-j+k+1, i=1..j)), j=1..k-1) + add( (-1)^(floor(j/k)+1)*(floor(j/k)-floor((j-1)/k))*(mul(j-i+1, i=1..k)), j=1..n): seq(a(n, 3), n=1..50); # Muniru A Asiru, Sep 30 2018
PROG
(PARI) Vec(3*x*(1 + x + 2*x^3 - 4*x^4 - 30*x^5 + x^6 - 5*x^7 + 24*x^8) / ((1 - x)*(1 + x)^4*(1 - x + x^2)^4) + O(x^50)) \\ Colin Barker, Oct 01 2018
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Wesley Ivan Hurt, Sep 30 2018
STATUS
approved